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X intercepts

  1. Jan 3, 2008 #1
    Ive got a function of f(x)=x^3+x and I need to use factors to show that the graph crosses the x axis once only.

    I just factorised it to x(x^2+1) which isn't very helpful, and if i divide everything by x and complete the square with x^2+1 i get a negative number and stuff....

    any help?
  2. jcsd
  3. Jan 3, 2008 #2


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    Oh yes it is!

    So, you have found that f(x)=x(x^2+1)

    What property, i.e, which equation must an x-intercept of the graph fulfill?
  4. Jan 3, 2008 #3
  5. Jan 3, 2008 #4


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    Good is that you factored the function. the x^2 + 1 factor has no "real" zeros; but the x factor has one real zero, being 0. (zero).

    Let me try again in case this helps.
    (x^2 + 1) = 0 for what real values of x? For NONE. We do not usually graph complex zeros in two dimensional cartesian plane (at least for our purposes here).

    When is x=0 (using the other factor)? when x=0, already shown.
  6. Jan 3, 2008 #5
    hi, thanks for the reply, it all makes sense now :)
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